Prove that $A-I_n$ is idempotent

Question

Let $A ∈ R^{n×n}$ be a nonzero singular matrix with the property that $A^2 = A$.

Show that $A − I_n$ is also idempotent.

I tried the following:

$(A - I_n)^2 = (A-I_n)(A-I_n) = A(A-I_n)-I_n(A-I_n) = A^2 - A - A + I_n = A^2 - 2A +I_n = A - 2A + I_n = I_n - A$

Unfortunately, to the best of my knowledge, that doesn't qualify as an answer, as the identity I must get is the following: $A − I_n$.

Since that does not equal to the final form that I got upon simplifying the square, I am sort of lost. Is there something that I did wrong in my steps?

Answer 1

Your computations are O.K. The statement above is wrong. If $A$ is idempotent, then $I_n-A$ is idempotent !

Answer 2

The correct statement is $I-A$ is idempotent.

If $$0 = A^2 - A = A(A - I)$$

then the minimal polynomial $\mu_A(x)$ divides $x(x-1)$.

If $\mu_A(x) = x$ then $A = 0$ and if $\mu_A(x) = x-1$ then $A = I$, both cases being trivial. Thus assume $\mu_A(x) = x(x-1)$.

Recall that for any matrix $B$ and scalar $\lambda$ we have $\mu_{B - \lambda I}(x) = \mu_B(x + \lambda)$. Also $\mu_{-B}(x) = (-1)^n\mu_{B}(-x)$.

Therefore $$\mu_{A - I}(x) = \mu_A(x+1) = (x+1)x \implies \mu_{I - A}(x) = x(x-1) = x^2 - x$$

We conclude $(I-A)^2 = I-A$.